Questions
A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :
detailed solution
Correct option is D
Initial energy stored in capacitor 2μFUi=122(V)2=V2Final voltage after switch 2 is ONVf=C1V1C1+C2=2V10=0.2VFinal energy in both the capacitorsUf=12C1+C2Vf2 =12102V102=0.2V2So, energy dissipated = V2-0.2V2V2 x 100=80%Talk to our academic expert!
Similar Questions
Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
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